I have SDE of the form $$dX_t=a\mathopen{}\left(X_t\right)dt+b\mathopen{}\left(X_t\right)dW_t,$$ where $W$ is Brownian motion. If the stationary distribution of $X$ exist is it equal to the limiting distribution? If not what kind of conditions would I need for them to be the same? I couldn't find any resources on this elsewhere. Thanks.
2026-04-14 11:26:04.1776165964
Conditions for limiting distribution to equal stationary distribution of SDE
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Partial answer: two situations where a distribution is stationary but not approached.
Periodicity: take $a(x)=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} x,b=0$. Then $X$ is periodic, any uniform distribution on any given circle is approached, but it is clear that if the initial distribution is a point mass anywhere except the origin, no distribution is approached.
Reducibility: if $a(x)$ and $b(x)$ have simultaneous zeros, then it can happen that there are multiple stationary distributions and so a given stationary distribution might not be approached. For example, $a(x)=-x^2(x-1)(x-2),b(x)=x(x-1)(x-2)$. This will have some distribution which is stationary and concentrated on $[0,1]$ and another concentrated on $[1,2]$. If the initial distribution is, say, uniform on $[0,2]$ then neither of these will be approached because $1$ is an impenetrable wall.
In the finite state space Markov chain context, these are the only things that can go wrong. When the state space becomes infinite, you can definitely have a problem with probability mass escaping "to infinity". If you prohibit that, too (by assuming that $b$ is not too huge and the drift drives you toward the origin if you stray too far from the origin), then you may be able to cook up some theorem.